Vector Analysis

In this lesson we will study the theoretical foundations of vector analysis, including vector operations, scalar and vector fields and differential operators. These concepts, together with the differential geometry of curves, are the prerequisite for studying line and surface integrals.

Vector Operations

Let’s begin with the space of ordered triples of real numbers $\mathbb{R}^3$. A point in this space is identified by three coordinates:

$$ P = (x, y, z) $$

A vector in $\mathbb{R}^3$, on the other hand, is represented by its three components:

$$ \mathbf{v} = (v_1, v_2, v_3) $$

Geometrically, a vector can represent a displacement, a velocity, a force or another quantity with both magnitude and direction.

Given two vectors $\mathbf{u}$, $\mathbf{v} \in \mathbb{K} \subseteq \mathbb{R}^3$, their dot product is a binary operation that produces a scalar:

$$ \mathbf{u} \cdot \mathbf{v} = (u_1 \cdot v_1) + (u_2 \cdot v_2) + (u_3 \cdot v_3) $$

Equivalently, if we know the magnitude of the two vectors and the angle $\theta$ between them, we can calculate their dot product using:

$$ \mathbf{u} \cdot \mathbf{v} = || \mathbf{u} || \cdot || \mathbf{v} || \cos \theta $$

For example, suppose that two vectors have magnitudes $4$ and $5$, respectively, and the angle between them is $60^\circ$. Their dot product is:

$$ \textbf{u} \cdot \textbf{v} = 4(5) \cos 60^\circ = 10 $$

If the dot product of two nonzero vectors is zero, the vectors are perpendicular.

The cross product, on the other hand, is a binary operation that takes two vectors in $\mathbb{R}^3$ and produces another vector. If we know the magnitudes of the two vectors and the angle between them, we can calculate the magnitude of their cross product using:

$$ || \textbf{u} \times \textbf{v} || = || \textbf{u} || \, || \textbf{v} || \sin \theta $$

where $\theta \in [0,\pi]$ is the angle between the vectors. For the previous example, the magnitude of the cross product is:

$$ || \textbf{u} \times \textbf{v} || = 4(5) \sin 60 = 10 \sqrt{3} $$

Notice that this gives the magnitude of the resulting vector, not the vector itself. To determine the cross product completely, we also need its direction.

Let $\textbf{u} = (u_1, u_2, u_3)$ and $\textbf{v} = (v_1, v_2, v_3)$ be two vectors. Their cross product can be written using the following determinant notation:

$$ \textbf{u} \times \textbf{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ u_1 & u_2 & u_3\\ v_1 & v_2 & v_3 \end{vmatrix} $$

where $\textbf{i}$, $\textbf{j}$ and $\textbf{k}$ are the unit vectors in the $x$, $y$ and $z$ directions, respectively. Expanding along the first row gives:

$$ \begin{aligned} \textbf{u} \times \textbf{v} &= \textbf{i} \begin{vmatrix} u_2 & u_3\\ v_2 & v_3 \end{vmatrix} - \textbf{j} \begin{vmatrix} u_1 & u_3\\ v_1 & v_3 \end{vmatrix} + \textbf{k} \begin{vmatrix} u_1 & u_2\\ v_1 & v_2 \end{vmatrix}\\[0.8em] &= (u_2 v_3 - u_3 v_2) \textbf{i} - (u_1 v_3 - u_3 v_1) \textbf{j} + (u_1 v_2 - u_2 v_1) \textbf{k} \end{aligned} $$

Equivalently, in component form:

$$ \textbf{u} \times \textbf{v} = \begin{pmatrix} u_2 v_3 - u_3 v_2\\ u_3 v_1 - u_1 v_3\\ u_1 v_2 - u_2 v_1 \end{pmatrix} $$

The resulting vector is perpendicular to both original vectors. Its magnitude equals the are of the parallelogram spanned by the two vectors and its direction is determined by the right-hand rule. If the two vectors are parallel, their cross product is the zero vector.

Let’s work through a complete example. Consider the following two vectors:

$$ \textbf{u} = (1, 2, 3), \quad \textbf{v} = (4, 5, 6) $$

Their cross product is:

$$ \begin{aligned} \textbf{u} \times \textbf{v} &= \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k}\\ 1 & 2 & 3\\ 4 & 5 & 6 \end{vmatrix}\\[1.5em] &= \textbf{i} (2 \cdot 6 - 3 \cdot 5) - \textbf{j} (1 \cdot 6 - 3 \cdot 4) + \textbf{k} (1 \cdot 5 - 2 \cdot 4)\\ &= -3\textbf{i} + 6\textbf{j} - 3\textbf{k} \end{aligned} $$

Therefore:

$$ \textbf{u} \times \textbf{v} = (-3, 6, -3) $$

We can now determine its magnitude:

$$ \begin{aligned} || \textbf{u} \times \textbf{v} || &= \sqrt{(-3)^2 + 6^2 + (-3)^2}\\ &= \sqrt{9 + 36 + 9}\\ &= \sqrt{54}\\ &= 3\sqrt{6} \approx \boxed{7.3484} \end{aligned} $$

We can verify this result using the sine formula:

$$ || \textbf{u} \times \textbf{v} || = || \textbf{u} || \, || \textbf{v} || \sin \theta $$

First, we calculate the dot product using:

$$ \textbf{u} \cdot \textbf{v} = ||\textbf{u}|| \, || \textbf{v} || \cos \theta $$

That is:

$$ \textbf{u} \cdot \textbf{v} = 1(4) + 2(5) + 3(6) = 4 + 10 + 18 = 32 $$

Next, calculate the magnitudes of the original vectors:

$$ || \textbf{u} || = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}, \quad || \textbf{v} || = \sqrt{4^2 + 5^2 + 6^2} = \sqrt{77} $$

Using the dot product formula:

$$ 32 = \sqrt{14} \sqrt{77} \cos \theta $$

Therefore, we can determine the angle:

$$ \cos \theta = \frac{32}{\sqrt{1078}} \implies \theta = \arccos \left( \frac{32}{\sqrt{1078}} \right) \approx 12.933154^\circ $$

We now have everything needed to calculate the magnitude of the cross product using the sine formula:

$$ \begin{aligned} || \textbf{u} \times \textbf{v} || &= \sqrt{14} \sqrt{77} \sin(12.933154^\circ)\\ &\approx \boxed{7.3484} \end{aligned} $$

which is the same result obtained directly from the components.

Scalar and Vector Fields

Let’s now define scalar and vector fields and highlight their main differences.

A scalar field is a function that assigns a real number to each point $(x, y, z) \in \mathbb{R}^3$ in its domain. In other words: $$ f : \mathbb{R}^3 \to \mathbb{R}, \quad f(x,y,z) \in \mathbb{R} $$

Examples of scalar fields include physical quantities such as the temperature at each point in a room or the density of a material at each point of a region.

A vector field is a function that assigns a three dimensional vector to each point $(x, y, z) \in \mathbb{R}^3$ in its domain. In other words: $$ \textbf{F} : \mathbb{R}^3 \to \mathbb{R}^3 $$ such that: $$ \textbf{F}(x, y, z) = (P(x,y,z), Q(x,y,z), R(x,y,z)) $$ Here, the three components of the vector are the values of the functions $P$, $Q$, and $R$ at the given point.

A vector field can represent a velocity, a force or any another quantity with both magnitude and direction.

These concepts will be relevant when studying line integrals. In a line integral of the first kind, we integrate a scalar field with respect to arc length. In a line integral of the second kind, instead, we integrate a vector field through its dot product with the curve’s tangent vector.

Let’s look at an example. Given the vector field: $$ F(x,y,z) = (x+y, y-z, xz) $$

and the point $A = (1,2,3)$, we want to evaluate $F(A)$:

$$ F(A) = (1 + 2, 2 - 3, 1\cdot 3) = (3, -1, 3) $$

Differential Operators

A differential operator is an operator that acts on a function or field by taking its derivatives. It can be used to study how a field varies in space. Let’s begin by introducing the following symbol:

$$ \nabla = \left( \frac{\partial}{\partial x}, \, \frac{\partial}{\partial y}, \, \frac{\partial}{\partial z} \right) $$

This is a formal differential operator which, depending on how it is used, produces the gradient, the divergence, or the curl.

Gradient

The gradient is a differential operator that acts on a scalar field and produces a vector field. It is defined by: $$ \nabla f = \left( \frac{\partial f}{\partial x}, \, \frac{\partial f}{\partial y}, \, \frac{\partial f}{\partial z} \right) $$

At a point where the gradient is nonzero, it points in the direction of the greatest local increase of the scalar field. Its magnitude gives the maximum rate of increase per unit distance. For instance, suppose that the temperature in a room is described by a scalar field $t(x, y, z)$. At any given point in the room, the gradient $\nabla t$ points in the direction in which the temperature increases most rapidly.

Let’s look at an example. Given the scalar field

$$ f(x,y,z) = x^2 y + 3z $$

we want to determine $\nabla f$. Let’s begin by calculating the partial derivatives:

$$ \frac{\partial f}{\partial x} = 2xy, \quad \frac{\partial f}{\partial y} = x^2, \quad \frac{\partial f}{\partial z} = 3 $$

We conclude that: $$ \nabla f = (2xy, x^2, 3) $$

Divergence

The divergence is a differential operator that takes a vector field and produces a scalar field. For a given $$ \textbf{F} = (P,Q,R) $$

it is defined by:

$$ \nabla \cdot \mathbf{F} = \operatorname{div}(\textbf{F}) = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} $$

The divergence measures the local net outward flow of a vector field per unit volume. A positive divergence indicates a local tendency to behave like a source, while a negative divergence indicates a local tendency to behave like a sink. Finally, a zero divergence indicates no local net source or sink.

For example, consider a vector field representing the velocity of water. Where the divergence is positive, more fluid flows outward from a sufficiently small region that flows inward; where it is negative, instead, more flows inward than outward. Naturally, the exact value depends on the field and the point being considered.

Let’s look at an example. Suppose we have the vector field:

$$ \textbf{F}(x,y,z) = (x^2y, 3yz^2, xz + 4z^2) $$

and we want to determine its divergence at the point$P(x,y,z) = (1,2,3)$. Let’s begin by calculating the three partial derivatives:

$$ \frac{\partial P}{\partial x} = 2xy, \quad \frac{\partial Q}{\partial y} = 3z^2, \quad \frac{\partial R}{\partial z} = x + 8z $$

Therefore:

$$ \nabla \cdot \textbf{F} = 2xy + 3z^2 + x + 8z $$

Finally, let’s evaluate this expression at the point $P$:

$$ \begin{aligned} \nabla \cdot \textbf{F}(P) &= 2(1)(2) + 3(3)^2 + 1 + 8(3)\\ &= 4 + 27 + 1 + 24\\ &= \boxed{56} \end{aligned} $$

Curl

The curl (also called the rotor) is a differential operator that takes a vector field and produces another vector field. It is defined by:

$$ \nabla \times \textbf{F} = \operatorname{rot}(\textbf{F}) = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) $$

Equivalently, we can use the following determinant notation as a mnemonic for the component formula:

$$ \nabla \times \mathbf{F} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\[0.5em] \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z}\\[0.8em] P & Q & R \end{vmatrix} $$

The curl measures the local rotational tendency of a vector field. For example, imagine a small paddle wheel placed in a flowing fluid. The curl describes the local axis and strength of the rotational tendency of the flow. Its direction is determined by the right-hand rule. In particular, for a rigid-body rotation, the angular velocity vector is one-half of the curl of the velocity field.

Let’s look at a practical example. Consider the following vector field:

$$ \mathbf{F}(x,y,z) = (x^2y, 3yz^2, xz + 4z^2) $$

and suppose we want to calculate its curl at the point $P(x,y,z) = (1,2,3)$. Let’s begin by evaluating the six partial derivatives appearing in the curl formula:

$$ \begin{matrix} \dfrac{\partial R}{\partial y} = 0, &\dfrac{\partial Q}{\partial z} = 6yz, & \dfrac{\partial P}{\partial z} = 0,\\[1.8em] \dfrac{\partial R}{\partial x} = z, &\dfrac{\partial Q}{\partial x} = 0, &\dfrac{\partial P}{\partial y} = x^2 \end{matrix} $$

Hence:

$$ \nabla \times \textbf{F} = (0 - 6yz, 0 - z, 0 - x^2) = (-6yz, -z, -x^2) $$

Finally, let’s evaluate this expression at the point $P$:

$$ \begin{aligned} \nabla \times \textbf{F}(P) &= (-6(2)(3), -3, -(1)^2)\\ &= \boxed{(-36, -3, -1)} \end{aligned} $$

Laplacian

The Laplacian (or Laplace operator) is a differential operator that takes a sufficiently smooth scalar field and produces another scalar field. In three dimensional Cartesian coordinates, it is defined by:

$$ \Delta f = \nabla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} $$

While, more generally, in $n$ dimensions:

$$ \Delta f = \sum_{k=1}^n \frac{\partial^2 f}{\partial x^2_k} $$

The Laplacian can also be expressed as the divergence of the gradient of a scalar field, that is:

$$ \Delta f = \nabla \cdot (\nabla f) $$